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Popular >

y^'=e^{y+t}

Solution

y^{^{'}} = e^{y + t}

Solution

y = - ln (- e^{t} - c_{1})

Solution steps

y^{'} (t) = e^{y + t}

Solve separable ODE:

y = - ln (- e^{t} - c_{1})

y = - ln (- e^{t} - c_{1})

Popular Examples

y*(dy)/(dx)=x*(x-e^x)/(e^{y^2)}

y \cdot \frac{d y}{d x} = x \cdot \frac{x - e ^{x}}{e ^{y^{2}}}

x^2y^'+1/x y= 2/x y^{-3}

x^{2} y^{^{'}} + \frac{1}{x} y = \frac{2}{x} y^{- 3}

(dy)/(dt)=0.07-y/(945)

\frac{d y}{d t} = 0.07 - \frac{y}{945}

(dy)/(dx)y=(x^2e^x)/(x^2+e^x)

\frac{d y}{d x} y = \frac{x ^{2} e ^{x}}{x ^{2} + e ^{x}}

x((dy)/(dx))-y=2x^2y

x (\frac{d y}{d x}) - y = 2 x^{2} y

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{y+t} ?
The solution for y^'=e^{y+t} is y=-ln(-e^t-c_{1})